The equation of the curve passing through the origin and satisfying the equation $(1 + {x^2})\frac{{dy}}{{dx}} + 2xy = 4{x^2}$ is

If $y=y(x)$ is the solution of the differential equation $\frac{dy}{dx}+\frac{4x}{x^2-1}y=\frac{x+2}{(x^2-1)^{5/2}}$ for $x > 1$,such that $y(2)=\frac{2}{9}\log_e(2+\sqrt{3})$ and $y(\sqrt{2})=\alpha\log_e(\sqrt{\alpha}+\beta)+\beta-\sqrt{\gamma}$,where $\alpha, \beta, \gamma \in N$,then $\alpha\beta\gamma$ is equal to $........$.

The general solution of the differential equation,$y' + y\phi'(x) - \phi(x)\phi'(x) = 0$,where $\phi(x)$ is a known function,is: (where $c$ is an arbitrary constant)

The solution of the differential equation $\cos x \, dy = y(\sin x - y) \, dx$ for $0 < x < \frac{\pi}{2}$ is:

Solve the differential equation $\left(\tan ^{-1} y-x\right) d y=\left(1+y^{2}\right) d x$.

Let $f:[2,5] \rightarrow R$ be a differentiable function and $\frac{f(5)}{f(2)}=1$. If there is a $c \in (2,5)$ such that $c f^{\prime}(c)=2 f(c)-2 c^3$, then $f(x)=$